a platinum resistance thermometer reads 0degrees and 100 degrees at ice point and boiling point of water respectively .the resistance of platinum wire varies with celcius temperature (theeta) as Rt=Ro(1+A(theeta)+B(theeta)square)

where A=3.8  into  (10raised to the power-3)

         B=5.6   into  (10 raised to power-7)

what would be the reading of this thermometer if placed in a liquid bath at 50 degree celcius.?

please tell me the priciples to help me do this sum and if possible even the solution

thank you

For paltinum resistance thermometer

 

R_t = R₀ (1 + A + B² )

 

where R₀ is resistance at 0 deg.

 

Now at 100 we have

 

R₁₀₀ = R₀ (1 + A + B² )

 

or R₁₀₀ = R₀ (1 + 3.8*10^-3 *100 + 5.6*10^-7 *100² )

 

or  R₁₀₀ = R₀ (1 + 0.38 + 5.6*10^-5)  this is the reading of the thermometer resistance at 100 deg.

 

 

Now at 50 deg. the reading of the platinum resistance thermometer is

 

R₅₀ = R₀ (1 + 3.8*10^-3 * 50  + 5.6*10^-7 * 50²) 

 

Here we substitute the value for R_o to obtain the value of resistance at 50 deg.

 

 

NOTE:   Resistance thermometers, also called resistance temperature detectors (RTDs), are temperature sensors that exploit the predictable change in electrical resistance of some materials with changing temperature. As they are almost invariably made of platinum, they are often called platinum resistance thermometers (PRTs). They are slowly replacing the use of thermocouples in many industrial applications below 600 °C.